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Fluid Dynamics

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  • 1. Ms. Arra C. Quitaneg 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 2. Pressure  Which of the two exerts greater pressure?  Why? A. m= 100 g B. m= 200 g 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 3. Pressure Which of the two exerts greater pressure? Why? A. m= 100 g B. m= 100 g 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 4. PRESSURE  Force per unit area  When a fluid is at rest, it exerts a force perpendicular to any surface in contact with it.  The force exerted by the fluid is due to the molecules colliding with its surroundings. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 5. Factors affecting Pressure  Force or Weight of the object  Area of support Pressure = Force/ Area Pressure is the ratio of the force to the area over which it is applied. Unit: 1 Pascal (Pa) = 1 N/m2 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 6. Do you want to try this? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 7. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 8. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 9. LIQUID PRESSURE  F = Ahg  P = F/ A  Liquid pressure = gh  =density of the liquid  g= acceleration due to gravity  h= depth 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 10. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 11. Fluid Pressure Pressure:  Units of measure: pounds per square inch (psi), atmospheres (atm), or torr (which is a millimeter of mercury).  The S.I. unit for pressure is the pascal, which is a Newton per square meter: 1 Pa = 1 N / m 2. Atmospheric pressure is at sea level is normally:  1 atm = 1.01 ·10 5 Pa = 760 torr = 14.7 psi.  At the deepest ocean trench the pressure is about 110 million Pascals. Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg 6/30/2013
  • 12. Fluid pressure Liquid pressure Atmospheric pressure  Fluid pressure in liquid varies with depth.  Due to the weight of the water column above the object submerged.  Due to the weight of the column of air above the surface.  Higher altitude – less atmospheric pressure 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 13. Which has greater liquid pressure? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 14. Liquid Pressure 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 15. Pressure of the Earth’s atmosphere. 1 atm = 1.013 x 105 Pa 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 16. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 17. Body moves from higher pressure to a lower pressure area. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 18. Water distribution  If a water tower is only a storage device, why is it so tall? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 19. WATER DISTRIBUTION  Water accelerates only if pressure is out of balance.  The deeper the water, the more weight there is overhead and the greater the pressure.  Water will not flow if there is no difference in pressure. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 20. Next time you take a shower… Oh! Thanks to the difference in pressure that I am able to take a shower! Physics works !!!!  6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 21. Think about these…  Consider an air –filled balloon weighted so that it is on the limit of sinking. That its over all density just equals that of water. Now, if you push it beneath the surface, it will a) Sink b) Return to the surface c) Stay at the dept to which it is pushed Explain your answer 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 22.  Farm silos have bands around them to provide sturdiness to the walls. Why are the bands closer together near the bottom of the silo than on top? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 23.  Explain how level hose works. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 24.  Diver Jeff designs a new scuba setup that is so profoundly simple he is surprised that no one has thought of this before. He has attached one end of a long garden hose to a large Styrofoam to keep the hose above the water level. He will breathe through the other end of the hose as he explores the depths. What is wrong with his simple design? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 25. Sample Problem: Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA Solution: Thus, the equivalent force is 240 N.  Problem 1:  A child wants to pump up a bicycle tire so that its pressure is 2.5 x 105 Pa above that of atmospheric pressure. If the child uses a pump with a circular piston 0.035 m in diameter, what force must the child exert? Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg A F P PAF NF 240 25 ) 2 035.0 )()(105.2( xF 6/30/2013
  • 26. Sample Problem: Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA  Problem 2:  In a classroom demonstration, a 73.5-kg physics professor lies on a “bed of nails.” The bed consists of a large number of evenly spaced, relatively sharp nails mounted in a board so that the points extend vertically outward from the board. While the professor is lying down, approximately 1900 nails make contact with his body. What is the average force exerted by each nail on the professor’s body? Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg6/30/2013
  • 27. Sample Problem: Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA  Problem 2:  In a classroom demonstration, a 73.5-kg physics professor lies on a “bed of nails.” The bed consists of a large number of evenly spaced, relatively sharp nails mounted in a board so that the points extend vertically outward from the board. While the professor is lying down, approximately 1900 nails make contact with his body. What is the average force exerted by each nail on the professor’s body?  Solution:  This gives about 0.379 N Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg A F P NF 379.0 nails S mkg F 1900 )8.9)(5.73( 2 A mg A W P 6/30/2013
  • 28. Sample Problem: Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA  Problem 3:  A swimming pool has the dimensions 15.0 m x 20.0 m. It is filled with water to a uniform depth of 8.00 m. The density of water = 1.00 × 103 kg/m3. Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg6/30/2013
  • 29. Sample Problem: Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA Natsci 24: Topic 1 (Hydrostatics) DLS-CSB-SMS-NATSCI AREA Solution: Thus, the equivalent pressure is 1.8 x 105 Pa  Problem 3:  A swimming pool has the dimensions 15.0 m x 20.0 m. It is filled with water to a uniform depth of 8.00 m. The density of water = 1.00 × 103 kg/m3. What is the total pressure exerted on the bottom of the swimming pool? Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg ghP PaxP 5 108.1 )8.9)(8)(100.1( 33 3 s mm m kg xP 6/30/2013
  • 30. Assignment  Bring the following by group: 2 identical drinking glasses 2 straws 1 pin ½ crosswise short bond paper 2 pingpong balls 1 yard length of a string Adhesive tape 2 empty aluminum cans (identical) hair dryer 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 31. Assignment  Clay  Aluminum foil  Coins  basin 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 32. Problem-solving:  The surface of the water in a storage tank is 30 m above a water faucet in the kitchen of a house, calculate the water pressure at the faucet. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 33. Pressure Pressure 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 34.  Estimate the pressure exerted on a floor by a) a 50 kg model standing momentarily on a single spiked heel (area = 0.05 cm2) and compare it b) to the pressure exerted by a 1500 kg elephant standing on one foot (area = 800 cm2). 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 35. Pascal’s principle  Blaise Pascal  Pressure applied to a confined fluid increases the pressure throughout by the same amount. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 36.  Any change in the pressure of an incompressible fluid is transmitted uniformly in all directions throughout the fluid.  Pressure is transferred undiminished in a fluid. P P 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 37. Pascal’s principle Application: Hydraulic Jack Changes in pressure are transmitted uniformly through a fluid and the pressure pushes outward in all directions. F A F A 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 38. Pascal’s principle oil A2 F1 A1 F2h1 h2 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 39.  A piston of cross-sectional area "a" is used in a hydraulic press to exert a small force of magnitude "f" on the enclosed liquid. A connecting pipe leads to a larger piston of cross-sectional area A. If the piston diameters are 3.80 cm and 53.0 cm, what force magnitude on the small piston will balance a 20.0 kN force on the large piston? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 40. Surface tension  Tendency of the liquid surface to contract.  Surface behaves as elastic film.  Caused by molecular attractions.  Soapy water has less surface tension than pure water. Spherical water droplet Water strider 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 41.  Why don’t boats made of steel SINK?  Is it possible for a material that is MORE DENSE than water to float? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 42. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 43.  A body immersed in water seems to weigh less than when it is in air.  Buoyant force- force exerted by liquid to an object immersed on it 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 44. Buoyant force 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 45. BUOYANT FORCE FB = F2 – F1 FB = g A (h2 – h1 ) FB = gV 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 46. Buoyant force FB = gV  =density of the liquid g=acceleration due to gravity V = volume of the object 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 47. Buoyancy  Accordingly, if the weight of the displaced water equals the weight of the object then it floats.  Buoyant force is now equal to the weight of the object.  This is known as the principle of flotation:  A floating object displaces a weight of fluid equal to its own weight. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 48. Archimedes principle The buoyant force acting on an object fully or partially submerged in a fluid is equal to the weight of the fluid displaced by the object. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 49. Floating of an object  Object floats if density of object is less than the density of the fluid.  An object is in equilibrium if the net force is zero. Buoyant force = Weight of the object 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 50. Fraction of the object submerged: 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 51. FLOATING RESTAURANTS 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 52.  A 15.0 kg solid gold statue is being raised from a sunken ship. What is the tension in the hoisting cable when the statue is at rest. A) completely immersed b) out of water  Density of gold= 19.3 x 103 kg/m3  Density of air = 1.2 kg/m3  Density of seawater- 1.025 x 103 kg/m3 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 53.  A wooden block of wood with a volume of 125 m3 floats on seawater. What fraction of object is submerged on seawater? Density of seawater- 1.025 x 103 kg/m3 Density of wood = 0.9 x 103 kg/m3 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 54. Seatwork  A block of steel measuring 2m on each side was accidentally dropped in seawater reaching a depth of 10 m. How much force is needed to lift the steel block if its density is 7.8 x 103 kg/m3 ? Density of seawater- 1.025 x 103 kg/m3  A 0.5 kg block of wood is floating in water. What is the magnitude of the buoyant force acting on the block? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 55.  A 70 kg ancient statue lies at the bottom of the sea. Its volume is 3.0 x 104 cm3. How much force is needed to lift it? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 56.  A small piston with a diameter of 0.5 m is used in a hydraulic lift. The larger piston has a diameter of 2m, how much force is needed to lift a car weighing 10 000 N ?  How much pressure is felt by woman standing on a heel with an area of 0.025 cm 2 , if her mass is 55 kg? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 57. The strongest aluminum boat!!!  Prepare 144 in2 aluminum foil.  Make your own boat design using the cut aluminum foil.  Prepare for the battle.  Which boat design will support more coins? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 58. POINTING SYSTEM NO . OF COINS POINTS 10 5 15 10 20 15 25 20 30 25 35 30 40 35 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 59. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 60. Activities……. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 61. When the speed of the fluid increases, pressure in the fluid decreases. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 62. Bernoulli's Principle  For Bernoulli's Principle to apply, the fluid is assumed to have these qualities: ◦ fluid flows smoothly ◦ fluid flows without any swirls (which are called "eddies") ◦ fluid flows everywhere through the pipe (which means there is no "flow separation") ◦ fluid has the same density everywhere (it is "incompressible" like water) Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg6/30/2013
  • 63. Bernoulli's Principle  Only true for a non-viscous fluid flowing at a constant height. It follows directly from the Bernoulli equation:  P + ½ v2 + gy = constant.  If y is a constant, then P + ½ v2 = constant.  This shows that if pressure increases, then speed decreases, and versa visa. Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg6/30/2013
  • 64. Airplane wing 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 65. Bernoulli’s principle Applications  Atmospheric pressure decreases in a strong wind, tornado or hurricane. Faster air LOWER PRESSURE HIGHER PRESSURE 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 66. The curved shape of an umbrella can be disadvantageous on a windy day! 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 67. Think…think…think…  Roofs of houses are sometimes blown off (or are they pushed off?) during a storm. Explain using Bernoulli’s principle.  An unvented building with airtight closed windows is in more danger of losing its roof than a well vented building. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 68.  Children are told to avoid standing too close to a rapidly moving train because they might get sucked under it. Is it possible? Why? 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 69. House roofing 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 70. Assignment  Which is the best roof design to prevent roof destruction due to difference in pressure above and below the roof. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 71. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg
  • 72. Reminders  Quiz on Monday  Part 1: Written exam  Part 2: Practical exam(floating restaurant design)  Bring all the materials you need for the design. Research on the weights of the objects you want to place in your floating restaurant. Practice computation. 6/30/2013 Prepared by: Ms. Paz Morales and Ms. Arra Quitaneg